Civil Engineering Reference
In-Depth Information
e
=
3.047
×
210000
×
11.9
2
/(
2
×
275
×
573.0
)
=
287.5
>
100
EC3-1-5, 6.5(3)
y
1
=
100
+
19.6
×
√
(
18.54
/
2
+
(
100
/
19.6
)
2
+
17.09
)
=
241.9
EC3-1-5, 6.5(3)
y
2
=
100
+
19.6
×
√
(
18.54
+
17.09
)
=
217.0
<
241.9 EC3-1-5, 6.5(3)
y
=
217.0mm.
EC3-1-5, 6.5(3)
217.0
×
11.9
×
275
1694
×
10
3
λ
F
=
=
0.648
>
0.5
EC3-1-5, 6.4(1)
χ
F
=
0.5
/
0.648
=
0.772 EC3-1-5, 6.2
L
eff
=
0.772
×
217.0
=
167.6mm EC3-1-5, 6.2
F
Rd
=
275
×
167.6
×
11.9
/
1.0N
=
548kN EC3-1-5, 6.2
<
558kN
=
R
Ed
, and so the assumed value of
(
s
s
+
c
)
=
100 mm
is inadequate.
5.12.16 Example 16 - designing a cantilever
Problem
. A3.0mlongcantileverhasthenominalimposedanddeadloads(which
include allowances for self weight) shown in Figure 5.44c. Design a suitable UB
section in S275 steel if the cantilever has sufficient bracing to prevent lateral
buckling.
Selecting a trial section
.
M
Ed
=
1.35
×
(
40
×
3
+
20
×
3
2
/
2
)
+
1.5
×
(
120
×
3
+
10
×
3
2
/
2
)
=
891 kNm.
Assume that
f
y
=
265 MPa and that the section is Class 2.
W
pl
,
y
≥
M
Ed
/
f
y
=
891
×
10
6
/
265mm
3
=
3362cm
3
6.2.5
Choose a 610
×
229 UB 125 with
W
pl
,
y
=
3676 cm
3
>
3362 cm
3
.
Checking the trial section
. The trial section must now be checked by classifying
thesectionandcheckingformoment,shear,momentandshear,andbearing.This
process is similar to that used in Section 5.12.15 for checking a simply supported
beam. If the section chosen fails any of the checks, then a new section must be
chosen.
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